CPhill

I'm stealing something here that I learned from heureka.....so, really....he should get the credit  !!!!

He noted that

n * n!  =

( [ n + 1 ]  - 1) * n! =

(n + 1)n! - n!  =

(n + 1)! - n!

So  we have the following

     1 * 1!           =    2!  -  1!

+   2 * 2!           =    3!  -  2!

+   3 * 3!           =    4!  -  3!

+   4 * 4!           =    5!  -  4!

+   .......

+   n * n!           =  (n + 1)!  - (n)! 

The terms in red  will "cancel"  and we will  be left with

Sum  = ( n + 1)!  - 1!  =

Sum  =  (n + 1)!  -  1

The nth Lucas number  can be derived as follows

L(n)  =  Phi^n  +  (-phi)^n

Where  

Phi  =   (1 + √5) / 2 )      and   -phi  =    -2 / ( 1 + √5)

So  

L(10) =    [ 1 +  [  (  (1 + √5) / 2 )^10  +   ( -2 / ( 1 + √5) )^10 ]    = 123

Also....if you know the Fibonacci Series....

L(n)  = F(n - 1)  + F(n + 1)

So

L(10)  =  F(9)  + F(11)   =   34  + 89  =   123

Thanks, hectictar  !!!

We have  the first few terms as

       2      8      20     40     70   

            6     12      20     30

                6       8       10

                     2       2

The Sum of Differences  produces 3 non-zero rows...so we will have a  cubic polynomial  representing the sum inside the parentheses

So we have that

a  +  b +  c   +  d   =  2

8a + 4b + 2c + d  =  8

27a + 9b + 3c + d  =  20

64a + 16b + 4c + d  =  40

Solving this system produces    a  = 1/3   b  =  1   c  = 2/3    d  = 0

So...the polynomial generating  the sums inside the parentheses  is just

(1/3)n^3 +  n^2  +  (2/3)n

So  3 times this is just

f (n)  =  n^3  + n^2  + 2n

Nvm..I read this incorrectly....hectictar spotted my error....I told her to post her solution   

I believe that hectictar's last answer is correct , as well

It appears that  f(x) might be  =  15(2)^x   since the points (0, 15)  and (1, 30) are on the graph

Rate  of change  on  -3 to -5  for   f(x)  =

[ 15(2)^(-3)  -  15 (2)^(-5) ] / [ -3  - - 5 ]  =  ( 15/8 - 15/32 ) / 2   ⇒  0.703

Rate of  change  on  -3 to - 5  for g(x)

[ 20(1.5)^(-3)  - 20(1.5)^(-5) ]  /  [ - 3 -  - 5 ]  =  (160/27 - 640/243 ) / 2  ⇒  1.646

So....."False" seems correct

Very nice, X²  !!!!

Yours  (and hectictar's )  visual presentations are always well done   (as opposed to being medium-rare......!!!  )

Here's my solution

AB  = 6, BC   = 10 and  AC  = 8

Call the point where the upper left vetex of the triangle meets the triangle  = D

Call the point where the upper right vertex of the triangle meets the triangle = F

Call the point where the bottom left vertex of the square meets the triangle = E

And we have three congruent triangles   ABC  ~ ADF  ~ EBD

Let  the side of the square  = x = DF    and let AD  = y

And we have that

DF / AD  = BC / AB

x / y   =   10/6     ⇒  y = (3/5)x

And 

BD / DE   = BC / AC

[ 6 - y ] / x  =  10 / 8        subbing for y, we have that

[6  -  (3/5)x ] / x  =  10 / 8  =  5/4

Cross-multiply

4 [ 6  - (3/5)x ]  = 5x

24  -  (12/5)x  =  5x

24  =  5x +  (12/5)x

24  =  [ 25  + 12]x / 5

120  = 37 x

x  ⇒  120 / 37    =   side of the square 

Wow!!... thanks, guest....that sure seems easier   !!!

QS...i answered this the other day....and hectictar  gave a more detailed  answer than mine today,  here :

https://web2.0calc.com/questions/pls-help_16#r1

Please check your old posts to see if any answers have been  given.....hectictar wasn't going to go through every page to see if something had been asked before  .....